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Assumption: Relationship In mode R(U, F), U=ABCDEG, F={B->D, DG->C, BD->E, AG->B, ADG->BC}
Find the minimum functional dependency set of F
Steps:
① Use the decomposition rule, Make the right part of any functional dependency in F contain only one attribute;
② Remove redundant functional dependencies:Remove it from F starting from the first functional dependency X→Y, and then Find the closure X of X in the remaining functional dependencies to see if X contains Y. If so, remove Until no redundant function dependencies are found;
③Remove the redundant attributes on the left side of each dependency. Check function dependencies one by one for dependencies on non-individual properties on the left side. For example, if XY→A is to be judged as redundant, is it equivalent to replacing XY→A with X→A? If A belongs to (X), then Y is a redundant attribute and can be removed.
Solution:
(1)
Determine whether the right side is the simplest, we get F={B- >D，DG->C,BD->E,AG->B,ADG->B,ADG->C}
(2)
① Assuming that B->D is redundant, remove B->D and get: G={DG->C,BD->E,AG->B,ADG-> B,ADG->C}B =B does not contain D, so it is not redundant and cannot be removed.
② Assuming that DG->C is redundant, remove DG->C and get: G={B->D,BD->E,AG->B,ADG-> ;B,ADG->C}(DG) =DG does not contain C, so it is not redundant and cannot be removed.
③ Assuming that BD->E is redundant, remove BD->E and get: G={B->D, DG->C, AG->B, ADG-> ;B,ADG->C}(BD) =BD does not contain E, so it is not redundant and cannot be removed.
④ Assuming that AG->B is redundant, remove AG->B and get: G={B->D, DG->C, BD->E, ADG-> ;B,ADG->C}(AG) =AG does not contain B, so it is not redundant and cannot be removed.
⑤ Assuming that ADG->B is redundant, remove ADG->B and get: G={B->D, DG->C, BD->E, AG-> ;B,ADG->C}(ADG) =ABCDEG contains B, so it is redundant and should be removed.
⑥Assuming that ADG->C is redundant, remove ADG->C and get: G={B->D, DG->C, BD->E, AG-> ;B}(ADG) =ABCDEG contains C, so it is redundant and should be removed.
In summary: F={B->D, DG->C, BD->E, AG->B}
(3)
① Assume that D->C is redundant, D =D does not contain C, so G cannot be removed.
② Assume that G->C is redundant, G =G does not contain C, so D cannot be removed.
③ Assume that B->E is redundant, B = BD does not contain E, so D cannot be removed.
④ Assume that D->E is redundant, D =D does not contain E, so B cannot be removed.
⑤Assume A->B is redundant, A =A does not include B, so G cannot be removed.
⑥Assume that G->B is redundant, G =G does not contain B, so A cannot be removed.
Therefore, Fm={B->D, DG->C, BD->E, AG->B}
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